8

WPF 的默认行为ContextMenu是在用户右键单击时显示它。我希望在ContextMenu用户左键单击时显示。看起来这应该是 上的一个简单属性ContextMenu，但事实并非如此。

我操纵它，以便我LeftMouseButtonDown在代码隐藏中处理事件，然后显示上下文菜单。

我在我的项目中使用 MVVM，这意味着我将DataTemplates 用于具有上下文菜单的项目。摆脱代码隐藏并找到一种使用 XAML 中的触发器或属性来显示上下文菜单的方法会更加优雅。

对此问题有任何想法或解决方案吗？

4

5 回答 5

13

我刚刚根据 HK1 的回答编写并测试了这个（您还可以阅读附加属性概述中的附加属性）：

public static class ContextMenuLeftClickBehavior
{
    public static bool GetIsLeftClickEnabled(DependencyObject obj)
    {
        return (bool)obj.GetValue(IsLeftClickEnabledProperty);
    }

    public static void SetIsLeftClickEnabled(DependencyObject obj, bool value)
    {
        obj.SetValue(IsLeftClickEnabledProperty, value);
    }

    public static readonly DependencyProperty IsLeftClickEnabledProperty = DependencyProperty.RegisterAttached(
        "IsLeftClickEnabled", 
        typeof(bool), 
        typeof(ContextMenuLeftClickBehavior), 
        new UIPropertyMetadata(false, OnIsLeftClickEnabledChanged));

    private static void OnIsLeftClickEnabledChanged(object sender, DependencyPropertyChangedEventArgs e)
    {
        var uiElement = sender as UIElement;

        if(uiElement != null) 
        {
            bool IsEnabled = e.NewValue is bool && (bool) e.NewValue;

            if(IsEnabled)
            {
                if(uiElement is ButtonBase)
                    ((ButtonBase)uiElement).Click += OnMouseLeftButtonUp;
                else
                    uiElement.MouseLeftButtonUp += OnMouseLeftButtonUp;
            }
            else
            {
                if(uiElement is ButtonBase)
                    ((ButtonBase)uiElement).Click -= OnMouseLeftButtonUp;
                else
                    uiElement.MouseLeftButtonUp -= OnMouseLeftButtonUp;
            }
        }
    }

    private static void OnMouseLeftButtonUp(object sender, RoutedEventArgs e)
    {
        Debug.Print("OnMouseLeftButtonUp");
        var fe = sender as FrameworkElement;
        if(fe != null)
        {
            // if we use binding in our context menu, then it's DataContext won't be set when we show the menu on left click
            // (it seems setting DataContext for ContextMenu is hardcoded in WPF when user right clicks on a control, although I'm not sure)
            // so we have to set up ContextMenu.DataContext manually here
            if (fe.ContextMenu.DataContext == null)
            {
                fe.ContextMenu.SetBinding(FrameworkElement.DataContextProperty, new Binding { Source = fe.DataContext });
            }

            fe.ContextMenu.IsOpen = true;
        }
    }

}

...

<Button Content="Do All" local:ContextMenuLeftClickBehavior.IsLeftClickEnabled="True" >
    <Button.ContextMenu>
        <ContextMenu>
            <MenuItem Header="Make everything awesome" />
            <MenuItem Header="Control the World" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

（注意 OnMouseLeftButtonUp() 方法中的注释）

于 2015-03-18T13:53:35.337 回答
9

我建议做的是制作一个带有附加 DependencyProperty 的新静态类。调用类 LeftClickContextMenu 和属性 Enabled（只是想法）。当您注册 DependencyProperty 时，添加一个 on changed 回调。然后在属性更改回调中，如果 Enabled 设置为 true，然后将处理程序添加到 LeftMouseButtonDown 事件并在那里执行您的操作。如果 Enabled 设置为 false，则删除处理程序。这允许您通过简单地在 xaml 中使用以下内容将其设置为任何属性。

<Border namespace:LeftClickContextMenu.Enabled="True" />

这种技术称为附加行为，您可以在此代码项目文章中了解更多信息：http: //www.codeproject.com/KB/WPF/AttachedBehaviors.aspx

于 2009-02-18T23:37:06.483 回答
4

虽然 Caleb 的答案是正确的，但它不包括工作代码。我使用 VB.NET 设置了一个示例（对不起），所以我将其发布在这里。

<Window x:Class="MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:AttachedBehaviorTest.AttachedBehaviorTest"
    Title="MainWindow" Height="350" Width="525">
    <Grid>
        <StackPanel>
            <TextBlock local:ContextMenuLeftClickBehavior.IsLeftClickEnabled="True">Some Text Goes Here
                <TextBlock.ContextMenu>
                    <ContextMenu>
                        <MenuItem Header="Test1" />
                    </ContextMenu>
                </TextBlock.ContextMenu>            
            </TextBlock>

        </StackPanel>
    </Grid>
</Window>

Namespace AttachedBehaviorTest

    Public NotInheritable Class ContextMenuLeftClickBehavior

        Private Sub New()
        End Sub

        Public Shared Function GetIsLeftClickEnabled(obj As DependencyObject) As Boolean
            Return CBool(obj.GetValue(IsLeftClickEnabled))
        End Function

        Public Shared Sub SetIsLeftClickEnabled(obj As DependencyObject, value As Boolean)
            obj.SetValue(IsLeftClickEnabled, value)
        End Sub

        Public Shared ReadOnly IsLeftClickEnabled As DependencyProperty = _
            DependencyProperty.RegisterAttached("IsLeftClickEnabled", GetType(Boolean), GetType(ContextMenuLeftClickBehavior), New UIPropertyMetadata(False, AddressOf OnIsLeftClickEnabled))

        Private Shared Sub OnIsLeftClickEnabled(sender As Object, e As DependencyPropertyChangedEventArgs)
            Dim fe As FrameworkElement = TryCast(sender, FrameworkElement)
            If fe IsNot Nothing Then
                Dim IsEnabled As Boolean = CBool(e.NewValue)
                If IsEnabled = True Then
                    AddHandler fe.MouseLeftButtonUp, AddressOf OnMouseLeftButtonUp
                    Debug.Print("Added Handlers")
                Else
                    RemoveHandler fe.MouseLeftButtonUp, AddressOf OnMouseLeftButtonUp
                    Debug.Print("RemovedHandlers")
                End If 
            End If
        End Sub

        Private Shared Sub OnMouseLeftButtonUp(sender As Object, e As RoutedEventArgs)
            Debug.Print("OnMouseLeftButtonUp")
            Dim fe As FrameworkElement = TryCast(sender, FrameworkElement)
            If fe IsNot Nothing Then
                'Next Line is Needed if Context Menu items are Data Bound
                'fe.ContextMenu.DataContext = fe.DataContext
                fe.ContextMenu.IsOpen = True
            End If
        End Sub

    End Class

End Namespace
于 2014-03-26T15:57:32.230 回答
1

这个答案与@nightcoder 的答案完全相同（感谢您的灵感！）。它使用 Blend 风格的行为，与附加属性相比，这是一种更现代的方法。

using System;
using System.Windows;
using System.Windows.Controls.Primitives;
using System.Windows.Data;
using System.Windows.Interactivity;

/// <summary>
/// Add this to any button menu allow a left click to open the context menu as well as the right.
/// </summary>
public class ContextMenuLeftClickBehavior : Behavior<ButtonBase>
{
    protected override void OnAttached()
    {
        base.OnAttached();

        this.AssociatedObject.Loaded += this.OnWindowLoaded;
        this.AssociatedObject.Unloaded += this.OnWindowUnloaded;
    }

    private void OnWindowLoaded(object sender, RoutedEventArgs e)
    {
        this.AssociatedObject.Click += OnMouseLeftButtonUp;
    }

    private void OnWindowUnloaded(object sender, RoutedEventArgs e)
    {
        this.AssociatedObject.Click -= OnMouseLeftButtonUp; // Cannot override OnDetached(), as this is not called on Dispose. Known issue in WPF.
    }

    private static void OnMouseLeftButtonUp(object sender, RoutedEventArgs e)
    {
        if (sender is ButtonBase fe && fe.ContextMenu != null)
        {
            if (fe.ContextMenu != null)
            {
                // If we use binding in our context menu, then it's DataContext won't be set when we show the menu on left click. It
                // seems setting DataContext for ContextMenu is hardcoded in WPF when user right clicks on a control? So we have to set
                // up ContextMenu.DataContext manually here.
                if (fe.ContextMenu?.DataContext == null)
                {
                    fe.ContextMenu?.SetBinding(FrameworkElement.DataContextProperty, new Binding { Source = fe.DataContext });
                }

                fe.ContextMenu.IsOpen = true;
            }
        }
    }
}

然后将行为添加到按钮：

<Button>
    <i:Interaction.Behaviors>
        <attachedProperties:ContextMenuLeftClickBehavior/>
    </i:Interaction.Behaviors>                                                
<Button>

像 Ellipse 或 Rectangle 这样的元素没有OnClick事件，这意味着没有什么能真正适用于任何交互的东西。因此，将所有内容包装在一个按钮中以获取该OnClick事件。也可以通过将鼠标光标更改为鼠标悬停时的手来暗示该区域是可点击的。

<Button.Style>
    <Style TargetType="{x:Type Button}">
        <Setter Property="Template">
            <Setter.Value>
                <ControlTemplate TargetType="{x:Type Button}">
                    <Ellipse Fill="{TemplateBinding Background}" Width="16" Height="16"/>
                </ControlTemplate>
            </Setter.Value>
        </Setter>
        <Style.Triggers>
            <Trigger Property="IsMouseOver" Value="True">
                <!--  Bind to custom color in ViewModel -->
                <Setter Property="Background" Value="{Binding CustomBrush}"/>
                <Setter Property="Cursor" Value="Hand"/>
            </Trigger>
        </Style.Triggers>
    </Style>
</Button.Style>
于 2019-03-04T13:07:00.990 回答
-1

忘记“唯一的xaml”的东西。当您将其包装到附加行为中时，可以很好地解决这个问题。

这是一种在左键单击时显示上下文菜单的方法：

Border在元素上创建一个新的左键处理程序：

<Border x:Name="Win"
        Width="40"
        Height="40"
        Background="Purple"
        MouseLeftButtonUp="UIElement_OnMouseLeftButtonUp">

然后添加：

private void UIElement_OnMouseLeftButtonUp(object sender, MouseButtonEventArgs e)
{
    e.Handled = true;

    var mouseDownEvent =
        new MouseButtonEventArgs(Mouse.PrimaryDevice,
            Environment.TickCount,
            MouseButton.Right)
        {
            RoutedEvent = Mouse.MouseUpEvent,
            Source = Win,
        };


    InputManager.Current.ProcessInput(mouseDownEvent);
}

它的作用是，它基本上将左键单击映射到右键单击。为了可重用性，您可以将其包装到附加的行为中。

于 2014-12-29T18:44:35.003 回答